Every group is a maximal subgroup of the free idempotent generated semigroup over a band

0.11 MB·English

by  Igor Dolinka

EVERY GROUP IS A MAXIMAL SUBGROUP OF THE FREE
IDEMPOTENT GENERATED SEMIGROUP OVER A BAND
IGORDOLINKAANDNIKRUSˇKUC
DedicatedtoStuartW.Margolisontheoccasionofhis60thbirthday
3
1 ABSTRACT. GivenanarbitrarygroupGweconstructasemigroupofidempotents
0 (band) B with the propertythat the freeidempotentgenerated semigroupover
G
2 B hasamaximalsubgroupisomorphictoG. IfGisfinitelypresentedthenB is
G G
n finite.Thisanswersseveralquestionsfromrecentpapersinthearea.
a
J

2
1
1. INTRODUCTION

] LetSbeasemigroup. ThesetE = E(S)ofallidempotentsofScarriesastructure
R
ofapartialalgebra,calledthebiordered setofS,byretainingtheproductsoftheso-
G
calledbasicpairs: thesearepairsofidempotents{e, f} suchthat{ef, fe}∩{e, f} 6=
.
h ∅. It should be noted that if ef ∈ {e, f} then fe is also an idempotent, possi-
t
a bly different from e, f and ef. Also, if S is an idempotent semigroup (i.e. a band)
m
then its biordered set is in general different from S itself, since not every pair is
[ necessarily basic. The term ‘biordered set’comes from an alternative (but equiva-

2 lent) approach, where one considers E(S) as a relational structure equipped with
v twopartialpre-orders;hereweshallnotpursuethisapproach,directinginsteadto
5 [4,5,6,11,16]forfurtherbackground.
2
The class of idempotent generated semigroups is of prime importance in semi-
2
1 group theory, with a host of natural examples, such as the semigroups of singular
. (non-bijective)transformationsofafiniteset(Howie[12])orsingularn×nmatrices
1
0 over a field (Erdos [7]). It is not difficult to show that the category of all idempo-
3 tent generatedsemigroupswith a fixed biordered set E has an initial object IG(E),
1
calledthefreeidempotentgeneratedsemigroupoverE(weshallalsosay‘overS’when
:
v E = E(S)). Thissemigroupisdefinedbythepresentation
i
X IG(E) = hE | e· f = ef ({e, f} isabasicpair)i.
r
a Here e· f stands for a word of length 2 in the free semigroup E+, while ef is the
elementofEtowhichtheproductequalsinS. Unsurprisingly,IG(E)playsacrucial
rule in understandingthe structureof semigroupswith a prescribedbiordered set
ofidempotents.
Forreasonsthatareintrinsictobasicstructuretheoryofsemigroups[11,13],this
in turn dependsupon the knowledgeof maximal subgroups of IG(E). It was con-
jectured for a long time that the maximal subgroups of IG(E) are always free; this
conjecture was widely circulated back in the1980s, and was explicitly recordedin
[15]. Theconjecturewasprovedinanumberofparticularcases,seee.g.[15,17,19].
In2009,Brittenham,MargolisandMeakin[1]disprovedtheconjecturebymeansof
2010MathematicsSubjectClassification. 20M05,20F05.
Keywordsandphrases. Freeidempotentgeneratedsemigroup,maximalsubgroup,band.
1
2 I.DOLINKAANDN.RUSˇKUC
anexplicit72-elementsemigroupSsuchthatIG(E(S))hasamaximalsubgroupiso-
morphictoZ⊕Z,thefreeabeliangroupofrank2. ThiswasfollowedbyGrayand
Rusˇkuc[9]whoprovedthatevery grouparisesasamaximal subgroupofIG(E(S))
forasuitablychosensemigroupS;ifthegroupinquestionisfinitelypresentedthen
afiniteS will suffice. Furtherensuingworksuchas[10,3,8]investigatesmaximal
subgroupsofIG(S) forsomespecificnaturalsemigroupsS,andthefirstauthor[2]
initiatesthestudyofIG(B),where Bisaband.
Theaimofthepresentnoteistoprovetheresultannouncedinthetitle:
Theorem 1. Let G be agroup. Then there exists aband B such that IG(B ) has a max-
G G
imal subgroup isomorphic to G. Furthermore, if G is finitely presented, then B can be
G
constructedtobefinite.
This single construction provides an alternative, simpler proof of all the main
results of [9] (Theorems 1–4), resolves [9, Problem 1] which asks whether every
finitely presented group is a maximal subgroup of IG(S) for some finite regular
semigroup S, and solves [2, Problem 2] which calls for a characterisation of maxi-
malsemigroupsoffreeidempotentgeneratedsemigroupsoverbands.
2. PRESENTATION FOR MAXIMAL SUBGROUPS
A general presentation for maximal subgroups of IG(S) in terms of parameters
that dependonly on the structure of S has been exhibited in [9, Theorem5]. Since
we are interested here only in the case of bands, we utilise the particular form of
thistheorem,deducedin[2,Corollary5].
Firstofall,recall[13,Theorem4.4.1]thatanyband Bdecomposesintoasemilat-
tice of rectangular bands, which are the D-classes of B. Thus a D-class D of S can
beviewedasan I×J ‘table’ofidempotentse (i ∈ I,j ∈ J),where{R : i ∈ I}and
ij i
{L : j ∈ J} are theR-andL-classesin D respectively. Fori,k ∈ I and j,l ∈ J we
j
refertothetuple(e ,e ,e ,e )asthe(i,k;j,l) square.
ij il kj kl
Supposenow wehave an element f ∈ B belongingto a D-class above D. From
thebasictheoryofbands(see,forexample,[13,Section4.4])weknowthat f induces
idempotentmappings σ : I → I,i 7→ σ(i), and τ : J → J, j 7→ (j)τ, such thatfor
alli ∈ I, j ∈ J wehave
feij = eσ(i),j, eijf = ei,(j)τ.
We say that the square (i,k;j,l) is singular induced by f if one of the following
holds:
(a) σ(i) = i,σ(k) = kand(j)τ = (l)τ ∈ {j,l}; or
(b) σ(i) = σ(k) ∈ {i,k} and(j)τ = j,(l)τ = l.
We talk of a left-right or up-down singular square depending on whether (a) or (b)
applies.
Withtheaboveconventionsthegeneralpresentationweneedisasfollows:
Proposition2([9,2]). ThemaximalsubgroupHofIG(B)containinge ∈ Dispresented
11
by
hf (i ∈ I, j ∈ J)| f = f = 1 (i ∈ I, j ∈ J), (1)
ij i1 1j
f−1f = f−1f ((i,k;j,l) asingularsquarein D)i. (2)
ij il kj kl
FREEIDEMPOTENTGENERATEDSEMIGROUPSOVERBANDS 3
3. CONSTRUCTION OF BG
LetG beanygroup. LetuschooseandfixapresentationhA | Ri for G inwhich
every relation has the form ab = c for some a,b,c ∈ A. It is clear that G has such
a presentation– for instance theCayley table would do. What is less obvious, but
nonethelessstilltrue,isthatifGisfinitelypresentedthenithasafinitepresentation
ofthisform. Onewayofseeingthisisasfollows: Arelation a ...a = b ...b can
1 k 1 l
be replaced by two relations of the form a ...a = c, b ...b = c, at the expense
1 k 1 l
of introducing a new generator c. Furthermore, the relation a ...a = c can be
1 k
replacedbyk−1relations a a = d , d a = d ,...,d a = d , d a = cof
1 2 2 2 3 3 k−2 k−1 k−1 k−1 k
thedesiredform,withnewgeneratorsd ,...,d .
2 k−1
Definesets
A = A∪{0}, A′ = {a′ : a ∈ A }, I = A ∪A′, J = A ∪{∞},
0 0 0 0 0 0
where0, ∞ and a′ (a ∈ A )are symbolsdistinctfromeach otherandthosealready
0
(l) (r) (l) (r)
in A. ConsiderthesemigroupT = T ×T ,whereT (respectivelyT )isthe
I J I J
semigroupofallmappings I → I (resp. J → J)writtenontheleft(resp. right). The
semigroupT hasauniqueminimal idealK consistingofall (σ,τ) withbothσ and
τ constant. Thisidealisnaturallyisomorphictotherectangularband I×J,andwe
willidentifythetwo. WewillvisualiseK asinFigure1.
0 ∞
0
0′
FIGURE 1. A visual representation of K = I × J, highlighting the
partition I = A ∪ A′, as well as the four distinguished rows and
0 0
columns.
Wenowdefineaset L ⊆ T . Allelements(σ,τ) ∈ Lwillhave
σ2 = σ, τ2 = τ, ker(σ) = {A ,A′}, im(τ) = A . (3)
0 0 0
Recall that ker(σ) is the equivalence on I defined by (i,i′) ∈ ker(σ) if and only
if σ(i) = σ(i′), and that it can be identified with the resulting partition of I into
equivalence classes. Therefore, each (σ,τ) will be uniquely determined by im(σ)
which must be a two-element setthat is a cross-sectionof {A ,A′}, and the value
0 0
(∞)τ ∈ A . Theelementsof L comein fourgroups: Z –theinitial pair; G, G –the
0
elementsarisingfromthegenerators A;R–theelementsarisingfromtherelations
R:
4 I.DOLINKAANDN.RUSˇKUC
Type Notation Indexing im(σ) (∞)τ
Z (σ ,τ ) – {0,0′} 0
0 0
G (σ ,τ ) a ∈ A {0,a′} a
a a
G (σ ,τ ) a ∈ A {a,a′} 0
a a
R (σr,τr) r = (ab,c) ∈ R {b,c′} a
TheseelementscanbevisualisedasshowninFigure2.
0 ∞ a ∞ 0 ∞ a ∞
0 0
b
a
0′
a′ a′
c′
Z : (σ0,τ0) G : (σa,τa) G : (σa,τa) R : (σr,τr)
FIGURE 2. The elements (σ0,τ0), (σa,τa), (σa,τa) (a ∈ A), (σr,τr)
(r = (ab,c) ∈ R) of L. For each (σ,τ) shadedare thetwo rows cor-
respondingtoim(σ) andonecolumn correspondingto (∞)τ. They
allhaveker(σ) = {A ,A′}andim(τ) = A .
0 0 0
Becauseker(σ) andim(τ)arethesameforall(σ,τ) ∈ Litfollowsthat Lisaleft
zero semigroup(i.e. xy = x for all x,y ∈ L). Furthermore,since K is an idealin T
(i.e. xy,yx ∈ K for all x ∈ K, y ∈ T ), the set B = K∪L is a subsemigroup of T .
G
We remark that, strictly speaking, B dependsnot only on G, but crucially on the
G
chosenpresentationforG.
4. PROOF OF THEOREM 1
WewillnowusethepresentationgiveninProposition2tocomputethemaximal
subgroup H of IG(B ) containing the idempotent e = (0,0) ∈ K. Relations (1) in
G 0
ourcontextread
f = f = 1(i ∈ I, j ∈ J). (4)
0j i0
Theremainingrelations(2)arisefromthesingularsquaresinducedbytheelements
of LactingonK. Eachup-downsingularsquareisofoneofthefollowingforms:
(a ,a ;c ,c ), (a′,a′,c ,c ) (a ,a ,c ,c ∈ A ).
1 2 1 2 1 2 1 2 1 2 1 2 0
Thesquare(a ,a ;c ,c )yieldstherelation
1 2 1 2
f−1 f = f−1 f (a ,a ,c ,c ∈ A ). (5)
a1,c1 a1,c2 a2,c1 a2,c2 1 2 1 2 0
Puttinga = c = 0, a = a, c = candusing(4)yields
1 1 2 2
f = 1(a,c ∈ A ); (6)
a,c 0
FREEIDEMPOTENTGENERATEDSEMIGROUPSOVERBANDS 5
clearly,alltheremainingrelations(5)areconsequencesof(6). Similarly,thesquares
(a′,a′,c ,c )yieldtherelations
1 2 1 2
fa′,c = f0′,c (a,c ∈ A0). (7)
(Notethatwedonotnecessarilyhave f0′,c = 1,andsocannotdeduce fa′,c = 1.)
Turningtotheleft-right singular squares,each (σ,τ) ∈ L inducespreciselyone.
Below we list respectively the squares introduced by (σ ,τ ) of type Z, (σ ,τ ) of
0 0 a a
type G, (σa,τa) of type G, and (σr,τr) of type R, together with the relations they
yield:
(0,0′;0,∞) : f0−,01f0,∞ = f0−′,10f0′,∞ (8)
(0,a′;a,∞) : f0−,a1f0,∞ = fa−′,1afa′,∞ (a ∈ A) (9)
(a,a′;0,∞) : fa−,01fa,∞ = fa−′,10fa′,∞ (a ∈ A) (10)
(b,c′;a,∞) : fb−,a1fb,∞ = fc−′,1afc′,∞ (r = (ab,c) ∈ R). (11)
Usingtherelations(4),(6),(7),wecantransform(8)–(11)into:
f0′,∞ = 1 (12)
fa′,∞ = f0′,a (a ∈ A) (13)
fa,∞ = fa′,∞ = f0′,a (a ∈ A) (14)
f0′,b = f0−′,1af0′,c (r = (ab,c) ∈ R). (15)
So, the group H is defined by the generators f (i ∈ I, j ∈ J) and relations (4),
i,j
(6),(7),(12)–(15). Therelations(4),(6),(7),(12)–(14)canbeusedsimplytoeliminate
allthegeneratorsexcept f0′,a (a ∈ A). Replacingeachsymbol f0′,a bythesymbol a,
theremainingrelations(15)become
ab = c(r = (ab,c) ∈ R).
∼
Inotherwords,weobtaintheoriginalpresentationfor G. Thisprovesthat H = G.
FinallynotethatifhA | Riisafinitepresentation,thesemigroupB isalsofinite,
G
with
|B | = (2|A|+2)(|A|+2)+1+2|A|+|R|,
G
andthiscompletestheproofofourtheorem.
5. AN EXAMPLE, TWO REMARKS AND AN OPEN PROBLEM
ItmaybeinstructivetofollowinaspecificexamplethesequenceofTietzetrans-
formations constituting the brunt of the above proof. Let us take G = Q , the
8
quaterniongroup,withthewellknownFibonacciF(2,3)presentation(see[14,Sec-
tion7.3]):
ha,b,c | ab = c, bc = a, ca = bi.
The dimension of K in this case is 8×5, and Proposition 2 gives a presentationin
termsof40generators. Thisis thensimplified byasequenceofgeneratorelimina-
tions,usingrelations(1),up-downsingularsquares,andleft-rightsingularsquares
induced by the elements of L of types Z, G, G. In the final step further singular
squaresarerevealed,givingbacktheoriginalpresentation.
If we record the original generators in a natural 8×5 grid, this process may be
encapsulatedasshowninFigure3.
6 I.DOLINKAANDN.RUSˇKUC
f0,0 f0,a f0,b f0,c f0,∞ 1 1 1 1 1
fa,0 fa,a fa,b fa,c fa,∞ 1 fa,a fa,b fa,c fa,∞
fb,0 fb,a fb,b fb,c fb,∞ 1 fb,a fb,b fb,c fb,∞
fc,0 fc,a fc,b fc,c fc,∞ −(→1) 1 fc,a fc,b fc,c fc,∞
f0′,0 f0′,a f0′,b f0′,c f0′,∞ 1 f0′,a f0′,b f0′,c f0′,∞
fa′,0 fa′,a fa′,b fa′,c fa′,∞ 1 fa′,a fa′,b fa′,c fa′,∞
fb′,0 fb′,a fb′,b fb′,c fb′,∞ 1 fb′,a fb′,b fb′,c fb′,∞
fc′,0 fc′,a fc′,b fc′,c fc′,∞ 1 fc′,a fc′,b fc′,c fc′,∞
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 fa,∞ 1 1 1 1 fa,∞ 1 1 1 1 fa,∞
1 1 1 1 fb,∞ 1 1 1 1 fb,∞ 1 1 1 1 fb,∞
−U−/→D 1 1 1 1 fc,∞ −L−/R−:→Z 1 1 1 1 fc,∞ −L−/R−:→G 1 1 1 1 fc,∞
1 a b c f0′,∞ 1 a b c 1 1 a b c 1
1 a b c fa′,∞ 1 a b c fa′,∞ 1 a b c a
1 a b c fb′,∞ 1 a b c fb′,∞ 1 a b c b
1 a b c fc′,∞ 1 a b c fc′,∞ 1 a b c c
1 1 1 1 1 1 1 1 1 1
1 1 1 1 a 1 1 1 1 a
1 1 1 1 b 1 1 1 1 b
L/R:G 1 1 1 1 c L/R:R 1 1 1 1 c
−−−→ −−−→
1 a b c 1 1 a b c 1
1 a b c a 1 a b c a
1 a b c b 1 a b c b
1 a b c c 1 a b c c
FIGURE 3. The sequence of Tietze transformations constituting the
proofofTheorem1.
Remark 3. Itis possibletodescribecompletelythestructureofthefreeidempotent
generated semigroup IG(B ). By known results (see e.g. [9, (IG1)–(IG4)]) IG(B )
G G
has precisely two regular D-classes. The ‘upper’ one is a left zero semigroup L
isomorphic to L (as all products in L are basic), while the ‘lower’ one K, the com-
pletelysimpleminimalideal,hasaReesmatrixrepresentationwithstructuregroup
G and (normalised) sandwich matrix (a−1), where (a ) is the |I|×|J| table that is
ji ij
theend-productofTietzetransformationsperformedintheproofofTheorem1(in
ourexamplethisisthelasttableinFigure3). WeclaimthatinfactIG(B ) = L∪K.
G
To confirm this, and see that the structure is completely determined, we need to
show how to write products ef and fe with e ∈ L, f ∈ K as products of idempo-
tentsfromKinIG(B). Fortheproductef notethatthereexistsg ∈ Ksuchthat fRg
and ge = g; both pairs {e,g} and {f,g} are critical and we have ef = egf = hf,
whereh = eg ∈ K. Theproduct fecanbetreatedsimilarly.
Remark 4. Associated to the biorder E of idempotents of a regular semigroup S
there is another free idempotent generated object RIG(E), the free regular idempo-
tent generated semigroup on E. It is the largest regular semigroup with the biorder
ofidempotentsE, anditspresentationcan beobtainedby addingfurtherrelations
to the defining presentation for IG(E). For definition and references we refer the
FREEIDEMPOTENTGENERATEDSEMIGROUPSOVERBANDS 7
reader to [9]. In particular, from (IG1)–(IG4), (RIG1), (RIG2) in [9] it follows that
IG(B ) = RIG(B ).
G G
OnewayofinterpretingRemarks3,4istosaythatthewordproblemforIG(B )
G
is decidable if and only if the word problem for G is decidable. It is the authors’
belief that the next stage in the ongoing exploration of free idempotent generated
semigroupsis preciselyan analysis ofthewordproblemfor IG(S). This at present
seemsadauntingtask,evenin thecasewhereS isfinite. Nonetheless,wepropose
thefollowingproblemwhichmayjustbewithinreachatthisstage:
Question 1. Let B be a finite band such that all maximal subgroups of IG(B) have
recursivelysolublewordproblems. IsthewordproblemofIG(B)necessarilyrecur-
sivelysoluble?
Acknowledgement. The research of the first author is supported by the Ministry
of Education, Science and Technological Development of the Republic of Serbia
throughGrantNo.174019,andbyagrant(Contract114–451–2675/2012) oftheSec-
retariat of Science and Technological Development of the Autonomous Province
of Vojvodina. Also, the first author gratefully acknowledgesthe hospitality of the
School of Mathematics and Statistics of the University of St Andrews, where this
researchwascarriedout.
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